Ask question

Ask question

All categories

2 years ago

12

Consider the reaction 2CuCl2 + 4K - 2Cul + 4KCI + 12. If 4 moles of CuCl2 react with 4 moles of KI, what is the limiting reactan

t?
a: CuCI2
b: KI
c: CuI
d: I2

1 answer:

Free_Kalibri [48]2 years ago

8 0

Haha i’m trying to do the same one i’ll make sure if i find out how too to get back to you!

You might be interested in

Protons Ha and Hb in the compound given are _________. homotopic enantiotopic diastereotopic mesotopic none of these

jek_recluse [69]

Protons Ha and Hb in the compound given are enantiotopic.

<h3>What are enantiotopic protons?</h3>

Enantiotopic protons can be defined as those protons that can be replaced by another groups like deuterium.

<h3>What is a compound?</h3>

A compound is a substance which contains two or more elements chemically combined together.

So therefore, Protons Ha and Hb in the compound given are enantiotopic.

Learn more about enantiotopic protons:

brainly.com/question/24106290

#SPJ1

8 0

1 year ago

Which statement is true? (2 points) Select one: a. Geothermal power plants use the renewable resource of heat from within the Ea

love history [14]

The correct answer is A, since it is the only sensible answer.

5 0

2 years ago

What terms refers to the process by which astronomers study distant objects by dissecting examining the spectrum of light an obj

mart [117]

Spectroscopy because it talks about the study of spectrum of light.

3 0

2 years ago

What is the oh of a solution with a ph of 9.8

Veseljchak [2.6K]

It would be 4.2, hope this helps.

4 0

2 years ago

A container is filled with helium gas. It has a volume of 2.25 liters and contains 9.00 moles of helium. How many moles of heliu

Darina [25.2K]

Answer: There are 7.4 moles of helium gas present in a 1.85 liter container at the same temperature and pressure.

Explanation:

Given: $V_{1}$ = 2.25 L, $n_{1}$ = 9.0 mol

$V_{2}$ = 1.85 L, $n_{2}$ = ?

Formula used to calculate the moles of helium are as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\$

Substitute the values into above formula as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{2.25 L}{9.0 mol} = \frac{1.85 L}{n_{2}}\\n_{2} = \frac{1.85 L \times 9.0 mol}{2.25 L}\\= 7.4 mol$

Thus, we can conclude that there are 7.4 moles of helium gas present in a 1.85 liter container at the same temperature and pressure.

7 0

1 year ago